# Variety of pairs of product-zero matrices

Here's an old qualifying exam question I got stuck on. Consider the variety $X$ of pairs of matrices $(A,B)$ satisfying $AB = BA = 0$ (with entries in some field). What are the irreducible components of $X$? According to the question, they all have dimension $n^2$.

A related question: are the irreducible components smooth away from their loci of intersection with other components?

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I would guess that the components are something like $$X_i=\{(A,B)\vert AB=BA=0, rank A \leq i, rank B \leq n-i\},$$ but I don't really know a proof. Can anybody assist? –  Boris Datsik Aug 12 '13 at 15:17

There is a paper of Gelfand and Ponomarev Indecomposable representations of the Lorentz group'' in wich such a variety is studied over an algebraicaly closed field.

I can not find the full text online, I am quite sure that one can download it from mathnet.ru but the site is not working today)

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Let $$\pi:X \rightarrow \mathrm{Mat}_n$$ be the projection $\pi(A,B)=A$. Stratify $\mathrm{Mat}_n$ by rank, with $V_i$ equal to the set of matrices of rank at most $i$. It is a fact that $V_i$ is irreducible of dimension $n^2-(n-i)^2$ (you can prove this e.g. by thinking about the correspondence consisting of pairs $(A,U)$ with $U \subseteq \mathrm{ker}(A)$ a subspace of dimension $n-i$).

Given $A \in V_i \setminus V_{i-1}$ a matrix of rank exactly $i$, we may choose a basis so that $A$ is in Jordan form. Let $J \subseteq \{1,2,\dots,n \}$ be the set of indices $j$ such that the $j$th column of $A$ is zero. Now the condition $AB=0$ is equivalent to each column of $B$ being in the kernel of $A$, or in other words that $B$ has non-zero elements only in rows indexed by $J$. On the other hand, the condition $BA=0$ is equivalent to the entries in $i$ of the columns of $B$ being zero (work by induction from left to right; exactly which columns depends on the Jordan structure). The following example illustrating the equation $0=BA$ in case $A$ has four Jordan blocks and kernel of dimension $2$, spanned by the fifth and seventh basis vectors, may help clarify how the argument proceeds: $$0=\left( \begin{matrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ a & b & c & d & e & f & g \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ h & i & j & k & l & m & n \end{matrix} \right) \left( \begin{matrix} 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} \right).$$

So the space of $B$ with $AB=0=BA$ is linear of dimension $(n-i)^2$. In other words, the fiber of $\pi$ over a point of $V_i \setminus V_{i-1}$ is a linear space of dimension $(n-i)^2$.

We have decomposed $X=X_0 \coprod X_1 \coprod \cdots \coprod X_{n-1} \coprod X_n$, where $X_i$ consists of pairs $(A,B)$ with $A$ of rank exactly $i$, and each $X_i$, having irreducible fibers of constant dimension $(n-i)^2$ over $V_i \setminus V_{i-1}$, is an irreducible locally closed subvariety of dimension $n^2-(n-i)^2+(n-i)^2=n^2$. The result follows from this (as does @Boris Datsik's observation about what the irreducible components are).

Also, it seems likely to me that the $X_i$'s are smooth since $V_i \setminus V_{i-1}$ is a homogeneous variety for the action of $\mathrm{GL}_n$ and $X_i$ seems to be a linear space bundle over it.

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